Question

prove that,-2, 4, and ½ are the zeros of the cubic polynomial p(x)=2x³-5x²-14x+8. Also verify the relation between the zeros and the cofficient

Answer 1

Hey dear !!!!!

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==> In the example ,

p(x) = 2x³ - 5x² - 14x + 8

We have to prove that -2 , 4 and 1/2 are the zeroes of given cubic polynomial or not ?

And also we have to verify the relation between the zeroes and the coefficient.

Lets first prove it ,

while taking -2 as zero of given cubic polynomial we get,

p(x) = 2x³ - 5x² - 14x +8

p(-2) = 2(-2)³ - 5(-2)² - 14(-2) +8

=2 * (-8) -5 *4 + 28 + 8

= -16 - 20 + 28 + 8

= -36 + 36

= 0

while taking 4 as zero of given cubic polynomial we get ,

p(4) = 2(4)³ -5(4)²- 14(4) + 8

= 2*64 -5* 16 -56 +8

= 128 - 80 -48

= 48 - 48

= 0

while taking 1/2 as zero of given cubic polynomial we get,

$p( \frac{1}{2} ) = 2( \frac{1}{2} ) {}^{3} - 5( \frac{1}{2} ) {}^{2} - 14( \frac{1}{2} ) + 8 \\ \\ = 2 \times \frac{1}{8} - 5 \times \frac{1}{4} - \frac{ 14}{2} + 8 \\ \\ = \frac{2}{8} - \frac{5}{4} - \frac{14}{2} + 8 \\ \\ = \frac{2 - 10 - 56}{8} + 8 \\ \\ = \frac{ - 64}{8} + 8 \\ \\ = \frac{ - 64 + 64}{8} \\ \\ = \frac{0}{8} \\ \\ = 0$

Therefore , we can say that -2,4 and 1/2 are the zeroes of the given cubic polynomial .

Now, relation between the zeroes and the coefficient,

Compare the given cubic polynomial with ax³ - bx² - cx + d we get ,

a = 4
b = -5
c = -14
d = 8

Also we assume that ,

α = -2
β = 4
γ = 1/2

We know that,

α + β + γ = -b/a

$- 2 + 4 + \frac{1}{2} = \frac{ - 4}{2} = \frac{ - b}{a}$

αβ + βγ + αγ = c/a

$- 2 \times 4 + \: 4 \times \frac{1}{2} + - 2 \times \frac{1}{2} = \frac{14}{2} = \frac{c}{a}$

αβγ = -d/a

$- 2 \times 4 \times \frac{1}{2} = \frac{ - 8}{2} = \frac{ - d}{a}$


The relation between the zeroes and the coefficient verified.

Thanks !!!☺

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